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Chapter 5: Problem 11

Give a recursive definition of \(P_{m}(n),\) the product of the integer \(m\) and the nonnegative integer \(n .\)

Short Answer

Expert verified
The recursive definition of \(P_{m}(n)\) is \(P_{m}(0) = 0\) and \(P_{m}(n) = P_{m}(n-1) + m\) for \(n > 0\).

Step by step solution

01

- Base Case

First, define the base case for the recursion. For any integer m, when n = 0, the product is 0. Hence, we can write: \[ P_{m}(0) = 0 \]
02

- Recursive Step

Now, define the recursive step. When n is positive, the product of m and n can be obtained by adding m to the product of m and (n-1). Therefore, for any positive integer n, the recursive definition can be written as: \[ P_{m}(n) = P_{m}(n-1) + m \]
03

- Combine Definitions

Combine the base case and the recursive step to form the complete recursive definition of P(m, n). The final recursive definition is: \[ P_{m}(n) = \begin{cases} 0 & \text{if } n = 0 \ P_{m}(n - 1) + m & \text{if } n > 0 \end{cases} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Base Case
A base case in a recursive definition is the simplest instance of the problem. It provides a stopping point for the recursion.
In our example of multiplying an integer \(m\) with a nonnegative integer \(n\), the base case is when \(n = 0\).
The product of any number and zero is always zero. Hence, our base case is:
\[ P_{m}(0) = 0 \]
This tells our recursion to stop when it reaches zero.
Recursive Step
The recursive step defines how the problem breaks down into smaller instances of itself.
For \( P_{m}(n) \), we need to find the product of \( m \) and \( n \).
If \( n \) is positive, we express the product \( P_{m}(n) \) in terms of \( P_{m}(n-1) \), which makes the problem smaller.
We can write:
\[ P_{m}(n) = P_{m}(n-1) + m \]
This means to get \( P_{m}(n) \), we add \( m \) to the product of \( m \) and \( n-1 \).
Combining this with our base case gives us a complete picture of how to calculate the product recursively.
Integer Multiplication
Integer multiplication is one of the most basic operations in mathematics.
Simply put, it is the process of repeated addition.
For example, multiplying \( 3 \) by \( 4 \) can be seen as adding \( 3 \), four times:
\[ 3 \times 4 = 3 + 3 + 3 + 3 = 12 \]
In our recursive definition, we achieve this by breaking down the multiplication into smaller, repeatable steps.
This approach aligns with the idea that multiplication is a repetitive addition process.
Discrete Mathematics
Discrete mathematics deals with distinct and separated values. It is fundamental in computer science and mathematical reasoning.
Recursive definitions are a big part of discrete math because they define functions, sequences, and operations in a step-by-step way.
The recursion for integer multiplication given above is a perfect example.
By breaking down the problem into a base case and a recursive step, we create a clear logical path that leads to the solution.
This structured approach is a cornerstone technique in discrete mathematics, helping to solve complex problems in a manageable way.

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Most popular questions from this chapter

Give a recursive algorithm for finding the string \(w^{i},\) the concatenation of \(i\) copies of \(w,\) when \(w\) is a bit string.

a) Give a recursive definition of the function \(m(s),\) which equals the smallest digit in a nonempty string of decimal digits. b) Use structural induction to prove that \(m(s t)=\) \(\min (m(s), m(t))\)

Let \(a_{1}, a_{2}, \ldots, a_{n}\) be positive real numbers. The arithmetic mean of these numbers is defined by $$ A=\left(a_{1}+a_{2}+\cdots+a_{n}\right) / n $$ and the geometric mean of these numbers is defined by $$ G=\left(a_{1} a_{2} \cdots a_{n}\right)^{1 / n} . $$ Use mathematical induction to prove that \(A \geq G\) .

Use the principle of mathematical induction to show that \(P(n)\) is true for \(n=b, b+1, b+2, \ldots,\) where \(b\) is an integer, if \(P(b)\) is true and the conditional statement \(P(k) \rightarrow\) \(P(k+1)\) is true for all integers \(k\) with \(k \geq b\) .

Give a recursive algorithm for tiling a \(2^{n} \times 2^{n}\) checkerboard with one square missing using right triominoes.
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